Introduction

Futures contracts are a common financial instrument used in crypto for hedging, speculation, or other purposes. Futures are linear, meaning their payoff (aka P/L) is linear w.r.t. the underlying asset (e.g. BTC in a BTC-USDT future).

A future position is defined by:

• Size $s$ (measured in the underlying asset, e.g. BTC)
• Entry price $p$ (measured in quote per underlying units, e.g. 100k USDT/BTC)

For example, a long position of 1 BTC entered at 100k would have $(s,p)=(1,100k)$. Conversely, a short position with same size and entry will have negative size, i.e. $(s,p)=(-1,100k)$.

Time Value

The value of a position $(s,p)$, given the underlying price $p_t$ at a time $t$ is

$$ V(t)=s(p_t-p) $$

For example, our 1BTC long @ 100k, if the current price is 110k, will have a value of 10k:

$$ V(t)=1\cdot(110k-100k)=10k $$

Conversely, the short will have negative value:

$$ V(t)=-1\cdot(110k-100k)=-10k $$

Linearity

With the value definition, we can easily see how it’s linear w.r.t. the underlying price $p_t$:

$$ \Delta=\frac{dV}{dp_t}=s $$

I.e., linearity means that the delta is constant. The change in value of the position is directly proportional to the underlying.

Equivalence

A nice thing of future contracts is their multiple dualities. Opening a long is equivalent to closing a short; closing a long equivalent to opening a short. In essence, only two operations are possible: going long and going short—buying and selling.

Further, multiple positions are always equivalent to a single position. Say for example you do these operations:

1. Long 2BTC @ 100k
2. Short 1BTC @ 110k
3. Long 1BTC @ 90k

What’s the value of the resulting position? Can it be described as a single future position, or do we require a composite equation? Turns out there’s an equivalent, single position. In this case, the resulting position will be $(s,p)=(2,90k)$.

Let’s prove it:

We have a set of positions $(s_1,p_1),…,(s_n,p_n)$, each with value $V_i(t)=s_i(p_t-p_i)$. Thus the total value is:

$$ V(t)=\sum_{i=1}^n V_i(t)=\sum_{i=1}^ns_i(p_t-p_i) $$

We can (1) split the sum and (2) take out $p_t$, since it’s a constant within the sum:

$$ V(t)=p_t\sum_{i=1}^ns_i-\sum_{i=1}^ns_ip_i $$

Now, we’ll call $s=\sum_{i=1}^ns_i$ the “total size”, fair enough:

$$ V(t)=p_ts-\sum_{i=1}^ns_ip_i $$

And, why not, let’s factor it out:

$$ V(t)=s\bigg(p_t-\frac{\sum_{i=1}^ns_ip_i}{s}\bigg) $$

What’s that thing on the right? Turns out it’s the “average entry price”—a weighted average of the entry prices (weighted by their size):

$$ p=\frac{\sum_{i=1}^ns_ip_i}{s} $$

Putting it all together:

$$ V(t)=s(p_t-p) $$

And what’s that? That’s the value of a future position of size $s$, entry price $p$. I.e., our set of positions $(s_1,p_1),…,(s_n,p_n)$ is equivalent to a single position $(s,p)$, defined as:

• $s=\sum_{i=1}^ns_i$, the total size
• $p={\sum_{i=1}^ns_ip_i}/{s}$, the average entry price

Let’s compute our example above:

• $s=2-1+1=2$
• $p=(2\cdot100k-1\cdot110k+1\cdot90k)/2=90k$

as promised.

Summary

We’ve seen some insightful things about future instruments:

1. A future position is entirely described by its size $s$ and entry price $p$
2. The value of a position is linear w.r.t. the underlying price
3. A set of future positions is equivalent to a single position with total size and average entry price